Solved CAT Questions (Algebra) - Set 4


  • Skadoooosh!!!


    Q1. (CAT 2000)
    For three distinct real numbers x, y and z, let
    f (x, y, z) = min(max(x, y), max(y, z), max(z, x))
    g(x, y, z) = max(min(x, y), min(y, z), min(z, x))
    h(x, y, z) = max(max(x, y), max(y, z), max(z, x))
    j(x, y, z) = min(min(x, y), min(y, z), min(z, x))
    m(x, y, z) = max(x, y, z)
    n(x, y, z) = min(x, y, z)

    Which of the following is necessarily greater than 1?
    (1) (h(x, y, z) – f(x, y, z))/j(x, y, z)
    (2) j(x, y, z)/h(x, y, z)
    (3) f(x, y, z)/g(x, y, z)
    (4) (f(x, y, z) + h(x, y, z) – g(x, y, z))/j(x, y, z)

    Which of the following expressions is necessarily equal to 1 ?
    (1) (f(x, y, z) – m(x, y, z))/(g(x, y, z) – h(x, y, z))
    (2) (m(x, y, z) – f(x, y, z))/(g(x, y, z) – n(x, y, z))
    (3) (j(x, y, z) – g(x, y, z))/h(x, y, z)
    (4) (f(x, y, z) – h(x, y, z))/f(x, y, z)

    Which of the following expressions is indeterminate?
    (1) (f(x, y, z) – h(x, y, z))/(g(x, y, z) – j(x, y, z))
    (2) (f(x, y, z) + h(x, y, z) + g(x, y, z) + j(x, y, z))/(j(x, y, z) + h(x, y, z) – m(x, y, z) – n(x, y, z))
    (3) (g(x, y, z) – j(x, y, z))/(f(x, y, z) – h(x, y, z))
    (4) (h(x, y, z) – f(x, y, z))/(n(x, y, z) – g(x, y, z))

    Solution: We need to work with options here. Don't get confused with all the functions and min/max representation. We will put it into a much more simpler way.

    For time being consider x > y > z ( say x = 2, y = 3 and z = 5)

    First function, f(2, 3, 5) = min(max(2, 3), max(3, 5), max(5, 2)) = min(3, 5, 5) = 3, which is the middle value. (we can get this by observation itself. I wrote the first one just as an explanation)

    Similarly g(2, 3, 5) - middle value (3)
    h(2, 3, 5) - max value (5)
    j(2, 3, 5) - min value (2)
    m(2, 3, 5) - max value (5)
    n(2, 3, 5) - min value (2)

    First question - Check in which of the given fractions Numerator > Denominator always. If it helps, you can put values and see the results. Only option 4 satisfies.
    Option 4 - ( f + h - g) / j = (3 + 5 - 3) / 2 > 1

    Second Question - Again check with options.
    Option 1 - (f - m)/(g - h) = (3 - 5)/(3 - 5) = 1.

    Third question - Option 2, Denominator is zero ( j + h - m - n = 2 + 5 - 5 - 2 = 0), hence indeterminate.

    Q2. (CAT 2000)
    Given below are two graphs made up of straight line segments shown as thick lines. In each case choose the answer as
    a. if f(x) = 3 f(–x)
    b. if f(x) = –f(–x)
    c. if f(x) = f(–x)
    d. if 3f(x) = 6f(–x), for x ≥ 0

    0_1489741207451_ar5.png

    Solution : c and d.
    first graph - symmetric graph with respect to y axis. This is a even function, f(x)=f(-x)
    second graph - this one is a symmetric graph too. But 2 * f(-1) = f(1). Answer will be the option where f(x)/f(-x) = 2, (option d)

    Q3. (CAT 2000)
    For real numbers x, y, let
    f(x, y) = Positive square-root of (x + y), if (x + y)^0.5 is real
    = (x + y)^2, otherwise
    g(x, y) = (x + y)^2, if (x + y)^0.5 is real
    = –(x + y), otherwise
    Which of the following expressions yields a positive value for every pair of non-zero real number (x, y)?
    (1) f(x, y) – g(x, y)
    (2) f(x, y) – (g(x, y))^2
    (3) g(x, y) – (f(x, y))^2
    (4) f(x, y) + g(x, y)

    Solution: f(x, y) is always positive.
    g(x, y) is always positive too.

    • (x + y)^2 always positive when (x + y)^0.5 is real
    • when (x + y)^0.5 is not real, x + y < 0 so - ( x + y ) is positive

    So f(x,y) + g(x,y) is always positive.

    Q4. (CAT 2000)
    For a real number x,
    let f(x) = 1/(1 + x), if x is non-negative
    = 1+ x, if x is negative
    f^n(x) = f(f^(n – 1)(x)), n = 2, 3, ....
    What is the value of the product, f(2) x f^2(2) x f^3(2) x f^4(2) x f^5(2) ?
    (1) 1/3
    (2) 3
    (3) 1/18
    (4) None of these

    f(2) = 1/3
    f^2(2) = f(f(2)) = f(1/3) = 1/(1 + 1/3) = 3/4
    f^3(2) = f(f^2(2)) = f(3/4) = 4/7
    f^4(2) = f(f^3(2)) = f(4/7) = 7/11
    f^5(2) = f(f^4(2)) = f(7/11) = 11/18
    Product = 1/3 x 3/4 x 4/7 x 7/11 x 11/18 = 1/18

    Q5. (CAT 2000)
    For all non-negative integers x and y, f(x, y) is defined as below
    f(0, y) = y + 1
    f(x + 1, 0) = f(x, 1)
    f(x + 1,y + 1) = f(x, f(x + 1, y))
    Then, what is the value of f(1, 2)?
    (1) Two
    (2) Four
    (3) Three
    (4) Cannot be determined

    Take third equation, f(x + 1, y + 1) = f(x, f(x + 1, y)
    We need to find f(1,2) so take x = 0 and y = 1.
    f(1, 2) = f(0, f(1, 1))
    = f(0, f(0, f (1, 0))
    = f(0, f(0, f(0, 1))
    = f(0, f(0, 2))
    = f(0, 3)
    = 4

    Q6. (CAT 2000)
    If the equation x^3– ax^2 + bx – a = 0 has three real roots, then it must be the case that,
    a) b = 1
    b) b ≠ 1
    c) a = 1
    d) a ≠ 1

    Make use of options.
    if b = 1, equation reduces to x^3 - ax^2 + x - a = 0
    x^2 (x - a) + (x - a) = 0
    (x - a) (x^2 + 1) = 0
    x = a or x^2 = -1
    Now we need three REAL solutions, which is not the case here.
    So b = 1 cannot be the case, so b should not be equal to one. (b # 1)

    Q7 - Q 10 (CAT 1999)
    Answer the questions based on the following information.
    In each of the following questions, a pair of graphs F(x) and F1(x) is given. These are composed of straight line segments, shown as solid lines, in the domain x is element of (-2, 2).
    a. if F1(x) = – F(x)
    b. if F1(x) = F(–x)
    c. if F1(x) = – F(–x)
    d. if none of the above is true

    Q7. 0_1490166148741_q7.png

    F(0) = 0 ; F1(0) = 0
    F(-2) = 2 ; F1(-2) = 2
    F(2) = 2 ; F1(2) = -2
    None of the above. (F(x) = | F(x) | )

    Q8. 0_1490166166027_q8.png

    F1(x) = F(-x), so option B

    Q9. 0_1490166177966_q9.png

    F1(x) = F(-x), so option B

    Q 10. 0_1490166190887_q10.png

    F1(x) = – F(–x). So Option C


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